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Question

If one root of the quadratic equation $$ { ax }^{ 2 }+bx+c=0$$ is equal to the $$n^{th}$$ power of the other, then show that 
$$ { \left( { ac }^{ n } \right)  }^{ \dfrac { 1 }{ n+1 }  }+{ \left( { a }^{ n }c \right)  }^{ \dfrac { 1 }{ n+1 }  }+b=0$$ 


Solution

Given that $$\beta = \alpha^n$$
Product of roots $$=\alpha \beta =\alpha { \alpha  }^{ n }={ \alpha  }^{ n+1 }=\dfrac { c }{ a } $$
$$\Rightarrow \alpha ={ \left( \dfrac { c }{ a }  \right)  }^{ \dfrac { 1 }{ n+1 }  }$$
Substituting $$\alpha$$ in $${ ax }^{ 2 }+bx+c=0$$
$$a{ \left( \dfrac { c }{ a }  \right)  }^{ \dfrac { 2 }{ n+1 }  }+b{ \left( \dfrac { c }{ a }  \right)  }^{ \dfrac { 1 }{ n+1 }  }+c=0$$
$${ \left( { ac }^{ n } \right)  }^{ \dfrac { 1 }{ n+1 }  }+{ \left( { a }^{ n }c \right)  }^{ \dfrac { 1 }{ n+1 }  }+b=0$$
Ans: 1

Mathematics

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